Using the normal distribution, it is found that 95.15% of students receive a merit scholarship did not receive enough to cover full tuition.
<h3>Normal Probability Distribution</h3>
The z-score of a measure X of a normally distributed variable with mean
and standard deviation
is given by:

- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
The mean and the standard deviation for the amounts are given as follows:

The proportion is the <u>p-value of Z when X = 4250</u>, hence:


Z = 1.66
Z = 1.66 has a p-value of 0.9515.
Hence 95.15% of students receive a merit scholarship did not receive enough to cover full tuition.
More can be learned about the normal distribution at brainly.com/question/15181104
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Answer:
So the height is 12
Step-by-step explanation:
Let W be the width
Let W- 4 be the height
W2 +(W-4)2 = 400
So: W2 -4W-192 =0
One uses the quadratic solution:
W = (4 + (16 + 4*192).5)/2 = 16
(13/20)=(x/152)
1976=20x
x=98.8
About 99 have dogs
30.50-14= 16.50
16.50 divided by 2.75= 6
The answer is 6
To find a percentage divide the first number by the second.

=

=1.00
Now that you have your number, move the decimal point over 2 to the right and add your percentage sign.
100.00%
Your answer is 100%